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3 years ago

How do iquantum tunel through a wall and how long will it take me?

Physics

1 answer:

Anon25 [30]3 years ago

5 0

Answer:

The phenomenon known as "tunneling" is one of the best-known predictions of quantum physics, because it so dramatically confounds our classical intuition for how objects ought to behave. If you create a narrow region of space that a particle would have to have a relatively high energy to enter, classical reasoning tells us that low-energy particles heading toward that region should reflect off the boundary with 100% probability. Instead, there is a tiny chance of finding those particles on the far side of the region, with no loss of energy. It's as if they simply evaded the "barrier" region by making a "tunnel" through it.

Explanation:

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The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

3 0

3 years ago

What property does matter have that energy does not'?

krok68 [10]

<span>The property that matter has that energy does not is that matter has size, shape and occupies space. Matter also has inertia. Energy does not have any of these.</span>

8 0

3 years ago

Read 2 more answers

at a drag race, the light turns green and 0.00125 hours later, a dragster is travelling 300 miles per hour. Calculate the accele

gtnhenbr [62]

<span>240,000 miles / hourÂ˛ Average acceleration can be calculated by dividing the change in speed by the elapsed time. Since the dragster's speed was 0 when the light turned green, the change in speed is simply 300 mph. Now, divide that by the time: 300 mph / 0.00125 hours = 240,000 miles / hourÂ˛ By the way, 0.00125 hours is just 4.5 seconds!</span>

5 0

4 years ago

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An apparatus like the one Cavendish used to ﬁnd G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.

Ber [7]

Answer:

The magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ .

Explanation:

Given that,

Mass of first lead ball, $m_1=5.2\ kg$

Mass of the other lead ball, $m_2=0.046\ kg$

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N$

So, the magnitude of gravitational force between two masses is $4.91\times 10^{-9}\ N$ . Hence, this is the required solution.

5 0

3 years ago

a basketball is tossed upwards with a speed of 5.0 m/s What is the maximum height reached by the basketball from its release poi

den301095 [7]

Answer:

1.275 m

Explanation:

Let the maximum height reached be h.

Here initial velocity, u = 5 m/s

Final velocity, V = 0

Use third equation of motion

V^2 = u^2 + 2 g h

0 = 25 - 2 × 9.8 × h

h = 25/19.6 = 1.275 m

3 0

4 years ago

Read 2 more answers